Question: Let $h$ be a vector-valued function defined by $h(t)=(2t^3+6,e^{2t})$. Find $h'(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $2t^2+2e^{t}$ (Choice B) B $(6t^2,2e^{2t})$ (Choice C) C $(2t^2,2e^{t})$ (Choice D) D $\left(\dfrac{t^4}2,\dfrac{e^{2t}}{2}\right)$
Solution: $h$ is a vector-valued function. This means it takes one number as an input $(t)$, but it outputs two numbers as a two-dimensional vector. Finding the derivative of a vector-valued function is pretty straightforward. Suppose a vector-valued function is defined as $u(t)=(v(t),w(t))$, then its derivative is the vector-valued function $u'(t)=(v'(t),w'(t))$. In other words, the derivative is found by differentiating each of the expressions in the function's output vector. Recall that $h(t)=(2t^3+6,e^{2t})$. Let's differentiate the first expression: $\begin{aligned}\dfrac{d}{dt}(2t^3+6)&=3\cdot2t^2 \\\\&=6t^2\end{aligned}$ Let's differentiate the second expression: $\dfrac{d}{dt}(e^{2t})=2e^{2t}$ Now let's put everything together: $\begin{aligned} h'(t)&=\left(\dfrac{d}{dt}(2t^3+6),\dfrac{d}{dt}(e^{2t})\right) \\\\ &=(6t^2,2e^{2t}) \end{aligned}$ In conclusion, $h'(t)=(6t^2,2e^{2t})$.